Corollary 2

Suppose that $\pi$ is $\nu$-proper; let $f$ be a mapping defined on $T$, with values in a Banach space or in $\overline{\mathbf{R}}$. For $f \circ \pi$ to be essentially $\nu$-integrable, it is necessary and sufficient that $gf$ be essentially $\mu$-integrable.

Taking into account Prop. 10, this follows immediately from Th. 1 of §5, No. 3 and Th. 1 of No. 2.